Given a string str containing lowercase characters, the task is to find the lexicographically largest sub-sequence of str.
Examples:
Input: str = “abc”
Output: c
All possible sub-sequences are “a”, “ab”, “ac”, “b”, “bc” and “c”
and “c” is the largest among them (lexicographically)Input: str = “geeksforgeeks”
Output: ss
Approach: Let mx be the lexicographically largest character in the string. Since we want the lexicographically largest sub-sequence we should include all occurrences of mx. Now after all the occurrences have been used, the same process can be repeated for the remaining string (i.e. sub-string after the last occurrence of mx) and so on until the there are no more characters left.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the lexicographically // largest sub-sequence of s string getSubSeq(string s, int n) { string res = ""; int cr = 0; while (cr < n) { // Get the max character from the string char mx = s[cr]; for (int i = cr + 1; i < n; i++) mx = max(mx, s[i]); int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) if (s[i] == mx) { res += s[i]; lst = i; } // Repeat the steps for the remaining string cr = lst + 1; } return res; } // Driver code int main() { string s = "geeksforgeeks"; int n = s.length(); cout << getSubSeq(s, n); } |
Java
// Java implementation of the approach class GFG { // Function to return the lexicographically // largest sub-sequence of s static String getSubSeq(String s, int n) { String res = ""; int cr = 0; while (cr < n) { // Get the max character from the String char mx = s.charAt(cr); for (int i = cr + 1; i < n; i++) { mx = (char) Math.max(mx, s.charAt(i)); } int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) { if (s.charAt(i) == mx) { res += s.charAt(i); lst = i; } } // Repeat the steps for // the remaining String cr = lst + 1; } return res; } // Driver code public static void main(String[] args) { String s = "geeksforgeeks"; int n = s.length(); System.out.println(getSubSeq(s, n)); } } // This code is contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach # Function to return the lexicographically # largest sub-sequence of s def getSubSeq(s, n): res = "" cr = 0 while (cr < n): # Get the max character from # the string mx = s[cr] for i in range(cr + 1, n): mx = max(mx, s[i]) lst = cr # Use all the occurrences of the # current maximum character for i in range(cr,n): if (s[i] == mx): res += s[i] lst = i # Repeat the steps for the # remaining string cr = lst + 1 return res # Driver code if __name__ == '__main__': s = "geeksforgeeks" n = len(s) print(getSubSeq(s, n)) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the lexicographically // largest sub-sequence of s static String getSubSeq(String s, int n) { String res = ""; int cr = 0; while (cr < n) { // Get the max character from // the String char mx = s[cr]; for (int i = cr + 1; i < n; i++) { mx = (char) Math.Max(mx, s[i]); } int lst = cr; // Use all the occurrences of the // current maximum character for (int i = cr; i < n; i++) { if (s[i] == mx) { res += s[i]; lst = i; } } // Repeat the steps for // the remaining String cr = lst + 1; } return res; } // Driver code public static void Main(String[] args) { String s = "geeksforgeeks"; int n = s.Length; Console.WriteLine(getSubSeq(s, n)); } } // This code is contributed by 29AjayKumar |
PHP
<?php // PHP implementation of the approach // Function to return the lexicographically // largest sub-sequence of s function getSubSeq($s, $n) { $res = ""; $cr = 0; while ($cr < $n) { // Get the max character from the string $mx = $s[$cr]; for ($i = $cr + 1; $i < $n; $i++) $mx = max($mx, $s[$i]); $lst = $cr; // Use all the occurrences of the // current maximum character for ($i = $cr; $i < $n; $i++) if ($s[$i] == $mx) { $res .= $s[$i]; $lst = $i; } // Repeat the steps for the // remaining string $cr = $lst + 1; } return $res; } // Driver code $s = "geeksforgeeks"; $n = strlen($s); echo getSubSeq($s, $n); // This code is contributed by // Akanksha Rai ?> |
ss
Time Complexity: O(N) where N is the length of the string.
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